Exercises — Spin — intrinsic angular momentum

Rule kya kehta hai: $m_s$ puri-number steps mein $-s$ se $+s$ tak chalta hai. Apply karo: $-\frac{1}{2}$ se $1$ upar jaane par $-\frac{1}{2},+\frac{1}{2}$ milta hai. Ruko (agla step $+\frac{3}{2}$, $+s$ se zyada ho jaata hai). Answer: $m_s=-\frac{1}{2}$ aur $m_s=+\frac{1}{2}$ — exactly do values. Exactly do kyun? Kyunki $2s+1=2(\frac{1}{2})+1=2$. Yahi wajah hai ki Stern–Gerlach do spots deta hai.

Rule: orientations ki sankhya $=2s+1$. Apply karo: $2(\frac{3}{2})+1=4$. Answer: 4 spots ($m_s=-\frac{3}{2},-\frac{1}{2},+\frac{1}{2},+\frac{3}{2}$ se).

Rule: $|\vec{S}|=\hslash \sqrt{s(s+1)}$. Apply karo: $\sqrt{\frac{1}{2}\cdot \frac{3}{2}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$. Answer: $|\vec{S}|=\frac{\sqrt{3}}{2}\hslash \approx 0.866\hslash$.

Cosine kyun use karein? $z$-axis ek right triangle ki ek side hai jiska hypotenuse $\vec{S}$ hai. $z$ ke along side shadow $S_z$ hai. "Adjacent over hypotenuse" exactly $\cos \theta$ hai — yahi woh tool hai jo shadow-length ko angle mein convert karta hai. $\cos \theta =\frac{S_z}{|\vec{S}|}=\frac{\frac{1}{2}\hslash }{\frac{\sqrt{3}}{2}\hslash }=\frac{1}{\sqrt{3}}.$ $\theta =\arccos \frac{1}{\sqrt{3}}\approx 54.7^{\circ }.$ Answer: $\theta \approx 54.7^{\circ }$. Yeh $0^{\circ }$ nahi hai — arrow tilts karta hai, hamesha.

$\cos \theta =\frac{-\frac{1}{2}\hslash }{\frac{\sqrt{3}}{2}\hslash }=-\frac{1}{\sqrt{3}}\Rightarrow \theta =\arccos \left(-\frac{1}{\sqrt{3}}\right)\approx 125.3^{\circ }.$ Negative sign kyun matter karta hai: negative cosine ka matlab hai arrow lower hemisphere ki taraf point karta hai ($90^{\circ }$ se aage ka angle). $125.3^{\circ }=180^{\circ }-54.7^{\circ }$, spin-up ka exact mirror. Dono cases cover ho gaye: up, $+z$ se $54.7^{\circ }$ tilt; down, $-z$ se $54.7^{\circ }$ tilt.

Rule: ${\mu }_z=-g_s\frac{e}{2m_e}{m}_s\hslash$. $g_s=2$ aur $|m_s|=\frac{1}{2}$ ke saath: $|{\mu }_z|=2\cdot \frac{e}{2m_e}\cdot \frac{1}{2}\hslash =\frac{e\hslash }{2m_e}={\mu }_B\approx 9.274\times 10^{-24}\text{J/T}.$ Answer: $|{\mu }_z|={\mu }_B\approx 9.27\times 10^{-24}\text{J/T}$. $g_s=2$ factor hi wajah hai ki spin double itna magnetic hai jitna naïve current-loop guess deta.

Setup: most-aligned state mein $S_z=s\hslash$ aur $|\vec{S}|=\hslash \sqrt{s(s+1)}$ hai. Toh $\cos \theta =\frac{s\hslash }{\hslash \sqrt{s(s+1)}}=\frac{s}{\sqrt{s(s+1)}}=\sqrt{\frac{s}{s+1}}.$ Limit: jab $s\to \infty$, $\frac{s}{s+1}\to 1$, toh $\cos \theta \to 1$ aur $\theta \to 0^{\circ }$. Interpret karo: bade $s$ (large systems) ke liye, arrow almost perfectly $z$ ke align ho sakta hai — quantum tilt negligible ho jaata hai aur hum classical picture recover karte hain. Chhote $s$ (jaise electron ka $\frac{1}{2}$) ke liye, tilt bada ($54.7^{\circ }$) aur irreducibly quantum hai. Yeh correspondence principle action mein hai. Edge sanity check: $s=\frac{1}{2}\Rightarrow \sqrt{\frac{1/2}{3/2}}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}$ ✔ L2.1 se match karta hai.

Do energies: ${\mu }_z=\mp {\mu }_B$ (up ke liye $-{\mu }_B\cdot (-1)$... carefully socho). $m_s=+\frac{1}{2}$ ke liye, ${\mu }_z=-{\mu }_B$, toh $U_{\uparrow }=-(-{\mu }_B)B_0=+{\mu }_B{B}_0$. $m_s=-\frac{1}{2}$ ke liye, ${\mu }_z=+{\mu }_B$, toh $U_{\downarrow }=-{\mu }_B{B}_0$. Gap: $\Delta U=U_{\uparrow }-U_{\downarrow }={\mu }_B{B}_0-(-{\mu }_B{B}_0)=2{\mu }_B{B}_0.$ Lines kyun split hoti hain: ek energy level do ban jaata hai (up aur down), $2{\mu }_B{B}_0$ se separate hoke. In se emit photons thodi different energies mein aate hain → ek spectral line close pair mein split ho jaati hai. Woh splitting $B_0$ ke saath scale hoti hai, jo Zeeman effect ka experimental signature hai.

(a) Force. $F_z={\mu }_z\frac{\partial B_z}{\partial z}$, ${\mu }_z=\pm {\mu }_B$ ke saath: $|F_z|=(9.274\times 10^{-24})(1.8\times 10^3)=1.669\times 10^{-20}\text{N}.$ Toh do forces hain $+1.67\times 10^{-20}$ N aur $-1.67\times 10^{-20}$ N — equal size, opposite sign → beam symmetrically do mein split ho jaati hai. (b) Velocity kick. Newton: $a=F_z/m$, phir $v_z=at$: $a=\frac{1.669\times 10^{-20}}{1.8\times 10^{-25}}=9.27\times 10^4{\text{m/s}}^2,$ $v_z=at=(9.27\times 10^4)(1.0\times 10^{-4})=9.27\text{m/s}.$ Answer: har atom $\approx 1.67\times 10^{-20}$ N se push hota hai aur $|v_z|\approx 9.3$ m/s ke saath nikalata hai (ek beam upar, ek neeche). $m_s$ ki discreteness do clean beams ke roop mein dikhti hai, smear nahi.

Pauli: do electrons jo $n,\ell ,m_{\ell }$ sab share karte hain, unka $m_s$ alag hona chahiye (dekho Quantum Numbers). Toh ek ka $m_s=+\frac{1}{2}$ aur doosre ka $m_s=-\frac{1}{2}$ hoga. Total shadow: $S_z^{\text{total}}=(+\frac{1}{2}+(-\frac{1}{2}))\hslash =0.$ Total moment: ${\mu }_z^{\text{total}}=(-{\mu }_B)+(+{\mu }_B)=0$. Answer: ek filled orbital spin-neutral hota hai: $S_z^{\text{total}}=0$ aur ${\mu }_z^{\text{total}}=0$. Isliye full shells koi net spin magnetism contribute nahi karte — do opposite arrows cancel ho jaate hain.

Setup: $L\approx m_{e}v{r}_e$ maano aur $L=\frac{1}{2}\hslash$ set karo. $v$ ke liye solve karo: $v\approx \frac{\hslash /2}{m_e{r}_e}=\frac{(1.055\times 10^{-34})/2}{(9.11\times 10^{-31})(2.82\times 10^{-15})}.$ Denominator compute karo: $9.11\times 10^{-31}\times 2.82\times 10^{-15}=2.569\times 10^{-45}$. Compute karo: $v\approx \frac{5.275\times 10^{-35}}{2.569\times 10^{-45}}\approx 2.05\times 10^{10}\text{m/s}$. Compare karo: $v/c\approx \frac{2.05\times 10^{10}}{3.0\times 10^8}\approx 68$. Conclusion: surface light ki speed se 68 baar tej move karegi — physically impossible. Isliye spin literal spatial rotation nahi hai. Yeh ek intrinsic quantum label hai, exactly jaisa parent note warn karta hai. (Rough $\sim 10^1$–$10^2$ overshoot standard textbook result hai; point enormous violation hai, last digit nahi.)

Sabse chhota angle sabse badi shadow se aata hai, $m_s=+s$ (L3.1 mein already nikala): $\cos {\theta }_{\min }=\sqrt{\frac{s}{s+1}}.$ Kabhi $0^{\circ }$ nahi: $\cos \theta =1$ ke liye $\frac{s}{s+1}=1$ chahiye, yaani $s+1=s$, kisi bhi finite $s$ ke liye impossible. Toh $\cos {\theta }_{\min }<1$ hamesha ⇒ ${\theta }_{\min }>0$ hamesha. Limit: $s\to \infty$ par, ${\theta }_{\min }\to 0^{\circ }$ (classical limit); electron ke liye $s=\frac{1}{2}$ par yeh maximum $54.7^{\circ }$ hai. Kyun (deep reason): agar arrow perfectly $z$ ke along hota, uski saari length $S_z$ mein hoti aur kuch bhi $S_x,S_y$ mein nahi — matlab $S_x=S_y=0$ exactly aur simultaneously jaane jaate. Commutator $[S_x,S_y]=i\hslash S_z\ne 0$ ise forbid karta hai. Figure s01 mein bacha hua length (jo side mein stick out karta hai) precisely woh uncertainty hai jo zaroor rehni chahiye. Toh tilt koi quirk nahi — yeh uncertainty principle ki geometry hai.

Projection $m$ use karta hai, size $\sqrt{s(s+1)}$ use karta hai ::: $S_z=m_s\hslash$ (shadow); $|\vec{S}|=\hslash \sqrt{s(s+1)}$ (full length) — doosra hamesha bada hota hai.